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\begin{document}
\title{Math 200b Winter 2022 Homework 6}
\date{Due 3/4/2020 by midnight on Gradescope, or hand in in class}
\maketitle



1.  In this problem you show that $G = \operatorname{Aut}(\mb{R}) = 1$.

\medskip

(a).  Show that every element of $\operatorname{Aut}(\mb{R})$
fixes $\mb{Q}$ pointwise; that is  $\operatorname{Aut}(\mb{R}) =  \Gal(\mb{R}/\mb{Q})$. 

\medskip

(b).  Let $\sigma \in G$.  Prove that $\sigma$ takes squares to squares and hence takes the set of positive numbers to itself.  Using this conclude that $a < b$ implies $\sigma(a) < \sigma(b)$.

\medskip

(c).  Prove that $\sigma$ is a continuous function.

\medskip

(d).  Prove that any continuous function $\mb{R} \to \mb{R}$ which fixes every element of $\mb{Q}$ is the identity, 
and so $\sigma = 1$.  

(Remark:  By constrast, it is possible to show that $\operatorname{Aut}(\mb{C})$ is uncountable!)

\bigskip

2.  Let $F \subseteq K$ be a finite degree field extension.  Suppose that 
$F \subseteq E_1 \subseteq K$ and $F \subseteq E_2 \subseteq K$ are intermediate 
fields and that $E_1/F$ and $E_2/F$ are normal extensions.  

\medskip

(a).  Show that  $(E_1 \cap E_2)/F$ is a normal extension.  

\medskip

(b).  Show that $(E_1E_2)/F$  is a normal extension.  The composite $E_1E_2$ is defined on the previous homework.

\bigskip

3.  Let $F \subseteq K$ be a finite degree Galois extension and let $F \subseteq E \subseteq K$ and $F \subseteq L \subseteq K$ be intermediate fields.  Show that there is an isomorphism of fields $\theta: E \to L$ with $\theta \vert_F = 1_F$ if and only if the subgroups $\Gal(K/L)$ and $\Gal(K/E)$ of $G = \Gal(K/F)$ are conjugate subgroups in $G$.

\bigskip

\bigskip

4.  Let $F$ be a field of characteristic $p > 0$, and let $\mb{F}_p$ be its prime subfield.  Let $K$ be the splitting field of $F$ of the polynomial $f(x) = x^p- x - a \in F[x]$. Let $\alpha \in K$ be a root of $f$ and assume that $\alpha \not \in F$.  Show that 

\medskip

(a). $\alpha + i$ is also a root of $f$, for all $i \in \mb{F}_p$.

\medskip

(b).  $K = F(\alpha)$.

\medskip

(c). $f$ is separable and irreducible in $F[x]$, and $K/F$ is Galois. (hint:  all roots of $f$ have minimal polynomials over $F$ of the same degree).

\medskip

(d).  For each $i \in \mb{F}_p$ there is an automorphism $\sigma_i \in G = \Gal(K/F)$ such that $\sigma_i(\alpha) = \alpha  + i$.  Moreover $G = \{ \sigma_i | i \in \mb{F}_p \}$ and $G$ is cyclic of order $p$.  

\bigskip


5.  Let $p_1, p_2, \dots, p_n$ be different prime numbers and let $E = \mb{Q}(\sqrt{p_1}, \dots, \sqrt{p_n})$ as a subfield of $\mb{C}$.

\medskip

(a).  Show that $E/\mb{Q}$ is Galois and that $\Gal(E/\mb{Q})$ is Elementary abelian of order $2^n$.  (Hint: show that the fields $\mb{Q}(\sqrt{k})$ are all different as $k$ runs over the $2^n -1$ different products of nonempty subsets of the set $\{p_1, p_2, \dots, p_n \}$.)

\medskip

(b).  Show that $E = \mb{Q}(\alpha)$, where $\alpha = \sqrt{p_1} + \dots + \sqrt{p_n}$.
(Hint:  determine how the $2^n$ elements of the Galois group $G$ act on the elements $\sqrt{p_1}, \dots, \sqrt{p_n}$.  Then show that the orbit of $\alpha$ under $G$ contains $2^n$ different elements.)

\medskip






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\end{document}
